TITLE: Closure of the product of subfunctors
QUESTION [5 upvotes]: Background:

Let $X: \textbf{CRing} \to \textbf{Set}$ be a presheaf on the category of affine schemes and $Z \subseteq X$ a subfunctor. One defines $Z$ to be closed if for every ring $A$ and every morphism $f: \text{Hom}(A , -) \to X$ the inverse image $f^{-1}(Z)$ is of the form $R \mapsto \{ \varphi : A \to R | \varphi(I) = 0 \}$ for some ideal $I \subseteq A$.
The intersection of subfunctors is defined naively, as is the closure (denoted by $\overline{Z}$) of a subfunctor $Z \subseteq X$ (it is the intersection of all closed subfunctors of $X$ containing $Z$).
If $Y$ is another presheaf, the product of $X$ and $Y$ is also defined naively.

Context: In section 1.14 of Jens Jantzen's great book "Representations of Algebraic Groups", the following is stated: If $X$ and $Y$ are presheaves which are schemes over a noetherian ring $k$ and $Z \subseteq X$ is a subscheme, and if $Z, X$ are algebraic and $Y$ is flat, then $\overline{Z \times Y} = \overline{Z} \times Y$. For the proof, he references Demazure-Gabriel I, section 2, 4.14 (although in my copy of Bell's translation this reference unfortunately doesn't exist).
Actual Question:
Is this true for general presheaves? I.e. if $X$ and $Y$ are presheaves and $Z \subseteq X$ is a subfunctor, is it true that $\overline{Z \times Y} = \overline{Z} \times Y$? I worry that it isn't because of the conditions in Jantzen stated above, but I haven't been able to decide either way. (Also side question: does anyone know the correct reference in the translation?)

REPLY [4 votes]: This is not true even for affine schemes. Let $k = \mathbb{Z}$, let $X = \operatorname{Spec} \mathbb{Z}$, let $Y = \operatorname{Spec} \mathbb{F}_p$, and let $Z \cong \operatorname{Spec} \mathbb{Z} [ p^{-1}]$. The closure of $Z$ in $X$ is $X$ itself, but $Z \times Y \cong \operatorname{Spec} \{ 0 \}$, which is already closed in $X \times Y$. Of course, $Y$ is not flat.